Need help for finding an integrating factor that makes a differential exact and solving it

When finding the integrating factor for $$e^x(x+1)dx+(ye^y-xe^x)dy=0$$ I used $$\frac{N_x-M_y}{M}=\frac{xe^x-e^x}{e^x(x+1)}$$ and got $$\frac{x-1}{x+1}$$. Then my integrating factor when solving for $$dy$$ is $$e^{(x-1)y/(x+1)}$$. When multiplying the original equation I got $$e^{(x-1)y/(x+1)}e^x(x+1)dx+e^{(x-1)y/(x+1)}(ye^y-xe^x)dy=0$$ but when finding $$M_y$$ for $$e^{(x-1)y/(x+1)}e^x(x+1)dx$$, I got $$(x-1)e^{((x-1)*y)/(x+1)+x)}.$$ For $$N_x$$ when using $$e^{(x-1)y/(x+1)}(ye^y-xe^x)dy$$, I got $$(((2*x*y+x+1)*e^y+(x-x^2)*e^x)*e^{(((x-1)*y)/(x+1)))/(x+1)}$$ which they are not the same. I tried integrating $$\frac{x-1}{x+1}$$ with $$dx$$ but I still don't get the same answer. So what did I do wrong?

• Please do not modify your initial question, since single questions are frowned upon on this site, they are closed for the most part. Instead your initial question showed your effort and precedence. Jan 30, 2022 at 21:50
• @Zaragosa You should roll back invalidating edits. Oct 11, 2022 at 16:51

It seems to me that you have made a mistake $$N_x=(ye^y-xe^x)_x=-xe^x-e^x=-e^x(x+1)$$, then you have $$\frac{N_x-M_y}{M}=-1$$. And therefore its integrating factor will be $$\mu(y)=e^{-y}$$. Multiplying the ODE by the integrating factor we have: $$e^{x-y}(x+1)dx+(y-xe^{x-y})dy=0$$ Thus having an exact ODE: Hence there is an $$F$$ such that $$F_x=M$$ and $$F_y=N$$. I think that from there you can continue on your own.